Puzzle for January 4, 2022  ( )

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Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 23 eq.2) C + E = F eq.3) A + B = F eq.4) B + D = A + E eq.5) C + E = D – E + F eq.6) D + E = A + C + F

A, B, C, D, E, and F each represent a one-digit non-negative integer.

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Hint #1


In eq.5, replace C + E with F (from eq.2): F = D – E + F In the above equation, subtract F from both sides, and add E to both sides: F – F + E = D – E + F – F + E which makes E = D


  

Hint #2


In eq.4, replace E with D: B + D = A + D Subtract D from each side of the above equation: B + D – D = A + D – D which makes B = A


  

Hint #3


In eq.3, substitute A for B: A + A = F which makes eq.3a) 2×A = F


  

Hint #4


Substitute D for E, and 2×A for F (from eq.3a) in eq.6: D + D = A + C + 2×A which becomes eq.6a) 2×D = 3×A + C


  

Hint #5


Substitute D for E, and 2×A for F (from eq.3a) in eq.2: C + D = 2×A Subtract C from each side of the equation above: C + D – C = 2×A – C which becomes eq.2a) D = 2×A – C


  

Hint #6


Substitute (2×A – C) for D (from eq.2a) into eq.6a: 2×(2×A – C) = 3×A + C which becomes 4×A – 2×C = 3×A + C In the equation above, add 2×C to both sides, and subtract 3×A from both sides: 4×A – 2×C + 2×C – 3×A = 3×A + C + 2×C – 3×A which makes A = 3×C and also makes B = A = 3×C


  

Hint #7


Substitute (3×C) for A in eq.2a: D = 2×(3×C) – C which becomes D = 6×C – C which makes D = 5×C and also makes E = D = 5×C


  

Hint #8


Substitute (3×C) for A in eq.3a: 2×(3×C) = F which makes 6×C = F


  

Solution

Substitute 3×C for A and B, 5×C for D and E, and 6×C for F in eq.1: 3×C + 3×C + C + 5×C + 5×C + 6×C = 23 which simplifies to 23×C = 23 Divide both sides of the above equation by 23: 23×C ÷ 23 = 23 ÷ 23 which means C = 1 making A = B = 3×C = 3 × 1 = 3 D = E = 5×C = 5 × 1 = 5 F = 6×C and ABCDEF = 331556